NEETClass 11thClass 12thClass 12th PlusJEEClass 11thClass 12thClass 12th PlusClass 6-10Class 6thClass 7thClass 8thClass 9thClass 10thOnline CoursesDistance LearningInternational OlympiadNEETClass 11thClass 12thClass 12th PlusJEE (Main+Advanced)Class 11thClass 12thClass 12th PlusJEE MainClass 11thClass 12thClass 12th PlusClass 6-10Class 6thClass 7thClass 8thClass 9thClass 10thKCET/MHT-CETKCETMHT-CETNEET2025202420232022JEE20262025202420232022Class 6-10JEE MainPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DatePercentile PredictorAnswer KeyCounsellingEligibilityExam PatternJEE MathsJEE ChemistryJEE PhysicsJEE AdvancedPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DateAnswer KeyEligibilityExam PatternRank PredictorNEETPrevious Year PapersSample PapersMock TestResultAnalysisSyllabusExam DateCollege PredictorAnswer KeyRank PredictorCounsellingEligibilityExam PatternBiologyNCERT SolutionsClass 6Class 7Class 8Class 9Class 10Class 11Class 12TextbooksCBSEClass 12Class 11Class 10Class 9Class 8Class 7Class 6SubjectsSyllabusNotesSample PapersQuestion PapersICSEClass 10Class 9Class 8Class 7Class 6State BoardBiharKarnatakaMadhya PradeshMaharashtraTamilnaduWest BengalUttar PradeshOlympiadMathsScienceEnglishSocial ScienceNSOIMONMTCASATInstant Online ScholarshipAIOT(NEET)TALLENTEXALLEN for SchoolsAbout ALLENBlogsNewsCareersRequest a call backBook a demo
  • Classroom Courses
  • NEW
  • ALLEN E-Store
Home
JEE Physics
Newton's Law of Viscosity

Frequently Asked Questions

Shear rate is the velocity gradient in a fluid, indicating how fast the velocity varies with the distance between the fluid layers.

Dynamic viscosity can be defined as the measure of the resistance of a fluid to flow under shear stress. The higher the value of dynamic viscosity, the thicker and more resistant the fluid.

Non-Newtonian fluids exhibit a change in viscosity as a function of shear rate, whereas Newtonian fluids have constant viscosity regardless of the rate of flow.

In gases, viscosity increases with pressure. In liquids, however, the effect of pressure on viscosity is usually negligible except in highly compressed liquids.

Join ALLEN!

(Session 2026 - 27)


Choose class
Choose your goal
Preferred Mode
Choose State

Related Articles:-

Difference Between Ammeter And Galvanometer

An ammeter and a galvanometer are two important tools used in electrical circuits to measure current, but they have different functions and operate in

Phase Angle: Definition

The phase angle is an important factor that shows where an oscillating object is at any given moment.

Bernoulli’s Principle: An Overview

A Swiss mathematician, Daniel Bernoulli, who formulated this theorem in the 18th century. The...

Buoyant Force

The buoyant force is the upward force exerted by any fluid (liquid or gas) on an object that is submerged in it or floating on its surface. It is responsible...

Hydraulic Machines

Hydraulic machines are very powerful tools that work on the principles of fluid mechanics to amplify force and execute a variety of tasks.

Understand Hydrostatic Paradox

Hydrostatic Paradox is that the pressure at any depth is independent of the size, shape, and aspect of the container, with the liquid's height being at par.

Master Photoelectric Effect

It is a phenomenon of ejecting electrons by falling light of a suitable frequency or wavelength on a metal. Ejected electrons are called

  • About
    • About us
    • Blog
    • Allen News
    • Privacy policy
    • Public notice
    • Careers
    • Dhoni Inspires NEET Aspirants
    • Dhoni Inspires JEE Aspirants
  • Help & Support
    • Refund policy
    • Transfer policy
    • Terms & Conditions
    • Contact us
  • Popular goals
    • NEET Coaching
    • JEE Coaching
    • 6th to 10th
  • Courses
    • Classroom Courses
    • Online Courses
    • Distance Learning
    • Online Test Series
    • International Olympiads Online Course
    • NEET Test Series
    • JEE Test Series
    • JEE Main Test Series
  • Centers
    • Kota
    • Bangalore
    • Indore
    • Delhi
    • More centres
  • Exam information
    • JEE Main
    • JEE Advanced
    • NEET UG
    • CBSE
    • NIOS
    • NCERT Solutions
    • Olympiad
    • NEET Mock Test
    • NEET Past Years Papers
    • NEET Sample Papers
    • NEET Answer Key 2026
    • NEET College Predictor 2026
    • NEET Rank Predictor 2026
    • NEET Cutoff
    • NEET Exam Analysis
    • NEET Revision Notes

ALLEN Career Institute Pvt. Ltd. © All Rights Reserved.

ISO

Newton's Law of Viscosity

Viscosity is a measurement of the resistance of a fluid to flow or deformation. It measures how much fluid resists the motion of sliding layers past one another. There is a mathematical approach under Newton's Law of Viscosity to describe this resistance as involving force and flow.

Law of Viscosity

1.0Statement and Explanation of Newton’s Law of Viscosity 

Newton's Law of Viscosity states that shear stress () is directly proportional to the related shear strain rate, which is defined as the velocity gradient between two layers. Another way to say this is the force that must be used to move one layer of fluid over another is proportional to the speed with which the velocity changes in the fluid. Mathematically, it can be written as: 

F=−ηdydu​

Statement of Newton's Law of Viscosity

Here:

  • F = shear stress (force per unit area (A) applied tangentially to the fluid),
  • η = dynamic viscosity of the fluid,
  • ​dudy = velocity gradient, which represents the rate at which the velocity of fluid changes with respect to the distance from the surface (i.e., the rate of shear strain).
  • The given formula is known as Newton’s law of viscosity Formula, which shows that for the Newtonian type of fluid, the relationship between shear stress and rate is linear. 
  • The negative sign indicates the direction of the shear force relative to the flow.

What is the Formula for Viscosity? 

The above-mentioned formula can also be used for deriving the formula for viscosity. It shows how the viscosity or internal resistance of fluids is related to the shear stress and the shear rate: 

F=dydu​τ​

  • The SI unit of dynamic viscosity - is pascal-seconds (Pa–s) while the CGS unit of viscosity is poise (P).
  • 1 poise (P) = 0.1 Pascal – second 

2.0Types of Fluids: 

Newtonian Fluids:

These fluids follow Newton's Law of Viscosity. The viscosity is constant and it is independent of the shear rate. The shear stress is always directly proportional to the shear rate. Examples of Newtonian fluids comprise water, air, and any other gas.

Types of Fluid

Non-Newtonian Fluids:

These fluids do not obey Newton's Law of Viscosity. Their viscosity varies with changes in the shear rate or external factors like temperature. Such compounds include ketchup, blood, paints, and some polymer solutions.

Non-Newtonian Fluids

3.0Solved Examples

Problem 1: Two fluids, A and B, are flowing between parallel plates. Fluid A has a dynamic viscosity of 0.4 Pa.s, and fluid B has a viscosity of 0.8 Pa.s. If the shear rate in fluid A is 10 s−1, calculate the shear rate for fluid B when the shear stress is the same in both fluids.

Solution: Let the Dynamic viscosity of fluid A A=0.4Pa.s

Let the Dynamic viscosity of fluid B B\eta_{B}=0.8Pa.s

The shear rate of fluid AdyduA​​=10s−1

According to the question, the shear stress is the same for both fluids. Hence,

FA=FB

ηA​dyduA​​=ηB​dyduB​​

0.4×10=0.8×dyduB​​

dyduB​​=0.80.4×10​=5s−1


Problem 2: A fluid flows through a pipe with a radius of 0.01 m. The pressure difference between the two ends of the pipe is 500 Pa. The velocity gradient in the pipe is 150 s−1. Find the viscosity of the fluid. 

Solution: Let the radius of the pipe be r = 0.01m 

Pressure difference be P = 500Pa

Velocity Gradient dydu​=150s−1

Here, we need shear stress for using Newton’s law of viscosity formula, which for pipe flow is given by: 

F=rΔP​

F=0.01500​=50000Pa

Now, using the formula of viscosity:

η=dydu​F​

η=150500000​=333.33Pa.s


Problem 3: A fluid flows between two parallel plates, with a distance of 0.02 m between them. The velocity at the lower plate is 0.3 m/s, and the velocity at the top plate is 0.5 m/s. If the dynamic viscosity of the fluid is 0.04 Pa.s, find the shear stress on the fluid.

Solution: Let the distance between plates by dy = 0.02m

Velocity at the lower plate u1,= 0.3m/s

Velocity at the top plate u2 = 0.5m/s

Dynamic viscosity of fluid = 0.04 Pa.s 

Velocity gradient can be calculated as:

dydu​=dyu2​−u1​​

=0.020.5−0.3​=0.020.2​

=10m/s2

Now, by using Newton’s law, let’s calculate shear stress:

F=ηdydu​

F=0.04×10=0.4Pa

Table of Contents


  • 1.0Statement and Explanation of Newton’s Law of Viscosity 
  • 1.1What is the Formula for Viscosity? 
  • 2.0Types of Fluids: 
  • 2.1Newtonian Fluids:
  • 2.2Non-Newtonian Fluids:
  • 3.0Solved Examples